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Date Submitted: 04/29/2014 11:42 PM
ASCI 309: Aerodynamics
Exercise I: Rectilinear Motion
Uniformly Accelerated Rectilinear Motion and Newton’s Law of Momentum
Equations to use (remember to keep track of units):
[pic]F =ma F-Force g-gravitational acceleration
[pic]F =(T-D-f)= (W/g)*a T-Thrust a-acceleration
Takeoff distance (s) = V2/2a D-Drag V-Velocity
s = ½ at2 f- friction t-time
KE = ½ mV2 W- Weight m-mass
KE-Kinetic Energy
s- distance
GIVEN INFORMATION: (Questions 1 to 5)
Gross Weight = 300,000 pounds
Average Drag = 15,000 pounds
Average Friction Force = 3,000 pounds
Average Thrust = 90,000 pounds
Lift Off Speed = 140 Knots True Airspeed (KTAS)
1. Compute the acceleration on the aircraft during the takeoff roll (ft / s 2).
w = mg
m = 300000 / 32
m = 9375slugs
a = F / m
a = (T – d – F) / m
a = (9000 – 15000 – 3000) / 9375
a = 7.68ft/s²
2. What would be the length of the takeoff run (ft)?
S = (V2 – V0²) / 2a
S = [[(140)(1.69)]² - 0] / 2 (7.68)
S = 3643ft
3. How long would it take until liftoff once the takeoff roll is started (s)?
V = V0 + at
140 = 0 + 7.68t
140 x 1.69 = 7.68t
t = (140 x 1.69) / 7.68
t = 30.8sec
4. Given the information shown above, determine how fast this airplane should be going when it passes the 2000-foot runway marker (2000 feet from the start of the takeoff roll). Express your answer in knots.
S = (V² – V0²) / 2a
2000 = (V² – 0) / 2 (7.68)
V² = (2000) (2) (7.68)
V = 175.27ft/s
V = 103.85knots
5. What is the Kinetic Energy of the aircraft after climbing out to 10,000 ft MSL and 250 KTAS?
V = 250 * 1.69
V = 422.5f/s
KE = ½mv²
KE = ½ * 9375 * (422.5)²
KE = 836748046.9ftlbs
KE = 8.3674 * 10^8ftlbs